Find out HP Required

How does the common controls engineer figure out how much HP is required for the application.

Say you have a conveyor pulling a belt, and you need to find a motor. How would I know what HP is required if it was all built in house?

How many pounds are lifted in a period of time? This would be the basic determining factor do power. Then add the efficiency (or rather losses) of the motor, gear box and friction losses in the conveyor etc.
There are some estimating guides available on the web from conveyor manufacturers.

Like this:http://test.roachconveyors.com/html/calculators/calculator/conveyorHP.php?ReportID=.

Most of the places I've been, motor sizing has been part of the mechanical design, or the motor vendors working with the mechanical team. If it had to be done by controls, we usually just had to go back, get all the info from mechanical, and then do it for them.

mk42 said:

Most of the places I've been, motor sizing has been part of the mechanical design, or the motor vendors working with the mechanical team. If it had to be done by controls, we usually just had to go back, get all the info from mechanical, and then do it for them.

Click to expand...

I think this is what I was expecting.

For sizing purposes, whether you are dealing with belts, sheaves, gearboxes, motors, or VFD's, you MUST understand what torque is required to do the work and how it changes over the speed range. Using the formula HP = T x RPM/5250, you find the HP. Be careful with this. On some loads, peak torque and peak speed do not occur at the same point. You have to size the components to cover both peak torque and peak speed and that can sometimes force you into a larger component than would normally be indicated by peak HP. It's always best to do your sizing using torque and speed, not HP.

Most i have seen is take the size of motor it would logically take and double or triple it.

There are a couple of things to take into consideration.
One is the pull out power or more correctly torque.
If the conveyor is stopped loaded this is an important consideration.
The constant running power is another. If you are lifting one ton 100 feet in a 60 seconds then plug these numbers in and calculate the horsepower for this and add the losses from friction etc.
The power losses are not as important usually as having the conveyor operating.
A second choice (depending on the power) would be a drive that would reduce the losses from a poor power factor.
Also you can run the motor at 2X the base motor speed and twice the gear reduction to give higher pull out torque to start the conveyor if it were to stop loaded.
It is not a lot of fun to have to shovel off the conveyor because it won't start loaded.

I usually come across the WAG method for motor sizing by customers, then if that doesn't work buy a bigger motor.

I have a customer that makes hoppers and feeders. He has a test conveyor that he loads up with parts he is setting up to handle (simulating the hopper dump) and applies a torque wrench to the belt axle to get an approximation.

This is fundamental physics.
1 hp = 33,000 ft-lbf / min
1 kW = 2.16*10^8 N-m / min = 60,000 m^2 - kg / min^3

Fundamentally you figure out how much you are lifting and how fast you are lifting it. Add a reasonable allowance for friction and you are done.

This site has some good info:http://www.engineeringtoolbox.com/conveyor-power-load-d_1560.html